Find the Moment about the ankle

[jweica]

Given that the tibia is 0.6kg and is 35 cm long. The force at the knee (Fk) is: (-200i + 346.41j)N
Find the moment about the ankle due to Fk

I found that: r: rx = (0.35cos60) and ry = (0.35sin60)

Using the sum of moments where: rxFy - ryFx = 0

[(0.35cos60)(346.41)] - [(0.35sin60)(-200)] = 0 k Nm

But I am wondering since the tibia weighs 0.6kg, wouldn't that be included in the question as a Fy component [(0.6)(9.81) = -5.886]?
Do I neglect the weight and only use the Fk?

((I attached a poorly drawn free body diagram))

 

[haruspex]

Given that the tibia is 0.6kg and is 35 cm long. The force at the knee (Fk) is: (-200i + 346.41j)N
Find the moment about the ankle due to Fk

I don't see anything there about 60 degrees. Where did that come from?

 

[Friend of Kalina]

I don't see anything there about 60 degrees. Where did that come from?

At a guess, it's from the dimensions of the force at the knee, where the angular force at the knee is broken into it's horizontal and vertical components of -200 and 346, which correspond to a 30/60/90 triangle having sides (2, 2##\sqrt (3)##, 4). It's unclear whether OP was given the angle and has already broken the forces into components, or was given the components and calculated the angle.

 

[haruspex]

At a guess, it's from the dimensions of the force at the knee, where the angular force at the knee is broken into it's horizontal and vertical components of -200 and 346, which correspond to a 30/60/90 triangle having sides (2, 2##\sqrt (3)##, 4). It's unclear whether OP was given the angle and has already broken the forces into components, or was given the components and calculated the angle.

Good point, but that means the 60 degrees is the angle of the applied force at the knee. The diagram shows it as the angle of the tibia. It seems unlikely to pose the question with the two at the same angle.

 

[Friend of Kalina]

Good point, but that means the 60 degrees is the angle of the applied force at the knee. The diagram shows it as the angle of the tibia. It seems unlikely to pose the question with the two at the same angle.

Given the diagram, I saw the problem as the force for someone turning with their leg at an angle to the ground and the foot/ankle pinned. The force is applied in line with the tibia (legs are don't bend laterally), with the leg at a 60 degree angle to the ground. Force in line with the leg along the tibia to the ankle is 400N, comprised of -200N horizontally and 346.41N vertically.

But you're right. Assuming that the components of Fk were given in the problem, if Fk is linearly along the tibia to the ankle, wouldn't that make it a radial force with zero torque? Nothing to solve, except the OP question of whether the tibia itself represents a moment. Much more likely that the tibia is vertical with the force Fk applied at the knee, in which case the tibia is vertical, provides no moment at the ankle, and can be disregarded. Although I'm not sure why you would then be concerned about the mass of the tibia, unless it was thrown into the problem as a red herring.

It would be nice to know if this was the original problem statement. It seems unlikely.